Answer :
To find the center and foci of the given ellipse, we need to rewrite the given equation in its standard form, which typically looks like:
[tex]\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the ellipse, and [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
Given the equation:
[tex]\[ 9x^2 + 16y^2 + 126x + 96y + 441 = 0 \][/tex]
We need to complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
### Completing the Square
First, we group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ 9x^2 + 126x + 16y^2 + 96y + 441 = 0 \][/tex]
#### For [tex]\(x\)[/tex]-terms:
[tex]\[ 9(x^2 + \frac{126}{9}x) = 9(x^2 + 14x) \][/tex]
To complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 14x \][/tex]
[tex]\[ = (x + 7)^2 - 49 \][/tex]
So,
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 \][/tex]
#### For [tex]\(y\)[/tex]-terms:
[tex]\[ 16(y^2 + \frac{96}{16}y) = 16(y^2 + 6y) \][/tex]
To complete the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 6y \][/tex]
[tex]\[ = (y + 3)^2 - 9 \][/tex]
So,
[tex]\[ 16(y + 3)^2 - 16 \cdot 9 \][/tex]
### Substitute Back:
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 + 16(y + 3)^2 - 16 \cdot 9 + 441 = 0 \][/tex]
Simplify the constants:
[tex]\[ 9(x + 7)^2 - 441 + 16(y + 3)^2 - 144 + 441 = 0 \][/tex]
Combine the constants:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 - 144 = 0 \][/tex]
### Standard Form:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 = 144 \][/tex]
Divide through by 144:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
Now we have the standard form of the ellipse equation:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
From this, we can identify:
- The center [tex]\((h, k)\)[/tex] of the ellipse is [tex]\((-7, -3)\)[/tex].
- The lengths of the semi-major axis [tex]\(a\)[/tex] and the semi-minor axis [tex]\(b\)[/tex]:
- [tex]\(a^2 = 16\)[/tex] so [tex]\(a = 4\)[/tex]
- [tex]\(b^2 = 9\)[/tex] so [tex]\(b = 3\)[/tex]
### Determining the Foci:
The distance [tex]\(c\)[/tex] from the center to each focus is found using:
[tex]\[ c = \sqrt{a^2 - b^2} \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ c = \sqrt{16 - 9} = \sqrt{7} \][/tex]
### Foci Coordinates:
The foci are along the [tex]\(x\)[/tex]-axis (horizontal) because [tex]\(a > b\)[/tex], so we add and subtract [tex]\(c\)[/tex] from the [tex]\(x\)[/tex]-coordinate of the center:
The foci are at:
[tex]\[ (-7 + \sqrt{7}, -3) \text{ and } (-7 - \sqrt{7}, -3) \][/tex]
### Answers:
- Center: [tex]\((-7, -3)\)[/tex]
- Foci: [tex]\((-7 \pm \sqrt{7}, -3)\)[/tex]
[tex]\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the ellipse, and [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the lengths of the semi-major and semi-minor axes, respectively.
Given the equation:
[tex]\[ 9x^2 + 16y^2 + 126x + 96y + 441 = 0 \][/tex]
We need to complete the square for the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms.
### Completing the Square
First, we group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ 9x^2 + 126x + 16y^2 + 96y + 441 = 0 \][/tex]
#### For [tex]\(x\)[/tex]-terms:
[tex]\[ 9(x^2 + \frac{126}{9}x) = 9(x^2 + 14x) \][/tex]
To complete the square for [tex]\(x\)[/tex]:
[tex]\[ x^2 + 14x \][/tex]
[tex]\[ = (x + 7)^2 - 49 \][/tex]
So,
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 \][/tex]
#### For [tex]\(y\)[/tex]-terms:
[tex]\[ 16(y^2 + \frac{96}{16}y) = 16(y^2 + 6y) \][/tex]
To complete the square for [tex]\(y\)[/tex]:
[tex]\[ y^2 + 6y \][/tex]
[tex]\[ = (y + 3)^2 - 9 \][/tex]
So,
[tex]\[ 16(y + 3)^2 - 16 \cdot 9 \][/tex]
### Substitute Back:
[tex]\[ 9(x + 7)^2 - 9 \cdot 49 + 16(y + 3)^2 - 16 \cdot 9 + 441 = 0 \][/tex]
Simplify the constants:
[tex]\[ 9(x + 7)^2 - 441 + 16(y + 3)^2 - 144 + 441 = 0 \][/tex]
Combine the constants:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 - 144 = 0 \][/tex]
### Standard Form:
[tex]\[ 9(x + 7)^2 + 16(y + 3)^2 = 144 \][/tex]
Divide through by 144:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
Now we have the standard form of the ellipse equation:
[tex]\[ \frac{(x + 7)^2}{16} + \frac{(y + 3)^2}{9} = 1 \][/tex]
From this, we can identify:
- The center [tex]\((h, k)\)[/tex] of the ellipse is [tex]\((-7, -3)\)[/tex].
- The lengths of the semi-major axis [tex]\(a\)[/tex] and the semi-minor axis [tex]\(b\)[/tex]:
- [tex]\(a^2 = 16\)[/tex] so [tex]\(a = 4\)[/tex]
- [tex]\(b^2 = 9\)[/tex] so [tex]\(b = 3\)[/tex]
### Determining the Foci:
The distance [tex]\(c\)[/tex] from the center to each focus is found using:
[tex]\[ c = \sqrt{a^2 - b^2} \][/tex]
Substitute [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ c = \sqrt{16 - 9} = \sqrt{7} \][/tex]
### Foci Coordinates:
The foci are along the [tex]\(x\)[/tex]-axis (horizontal) because [tex]\(a > b\)[/tex], so we add and subtract [tex]\(c\)[/tex] from the [tex]\(x\)[/tex]-coordinate of the center:
The foci are at:
[tex]\[ (-7 + \sqrt{7}, -3) \text{ and } (-7 - \sqrt{7}, -3) \][/tex]
### Answers:
- Center: [tex]\((-7, -3)\)[/tex]
- Foci: [tex]\((-7 \pm \sqrt{7}, -3)\)[/tex]
